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Standard XII

Chemistry

Gay Lussac's Law, Avagadro's Law

One mole of m...

Question

One mole of mixture of $C H_{4}$ and air (containing 80% $N_{2}$ 20% $O_{2}$ by volume) of a composition such that when underwent combustion gave maximum heat (assume combustion of only $C H_{4}$ . Then which of the statements are correct, regarding composition of initial mixture. (X presents mole fraction)

XCH₄ =O2 =

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XCH₄ = =

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XCH₄=1/6XO₂ =X_N2 =2/3

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Data insufficient

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Solution

The correct option is A
XCH₄ =O2 =

$C H_{4}$ + $O_{2}$ $⟶$ $C O_{2}$ +2 $H_{2} O$

Xmole 1-x mole

$n_{(0_{2})}$ = 0.2(1-x)

For complete combustion

$\frac{n_{C H_{4}}}{1}$ = $\frac{n o_{2}}{1}$

$\frac{x}{1}$ = $\frac{0.2 (1 - x)}{2}$

2x =0.2-0.2x

2-2x =0.2

X = $\frac{x}{11}$

$X_{C H_{4}}$ = $\frac{1}{11}$

$X_{O_{2}}$ = $\frac{2}{11}$ $X_{(N_{2})}$ = $\frac{8}{11}$

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One mole of mixture ofCH4 and air (containing 80%N2 20%O2 by volume) of a composition such that when underwent combustion gave maximum heat (assume combustion of only CH4. Then which of the statements are correct, regarding composition of initial mixture. (X presents mole fraction)

The correct option is A XCH4 =O2 = CH4+O2⟶CO2+2H2O Xmole 1-x mole n(02)= 0.2(1-x) For complete combustion nCH41 =no21 x1 =0.2(1−x)2 2x =0.2-0.2x 2-2x =0.2 X =x11 XCH4 =111 XO2 =211X(N2)=811